Mathematics
In the given figure, O is the centre of the circle. If ∠AOD = 140° and ∠CAB = 50°, calculate :
(i) ∠EDB
(ii) ∠EBD
Circles
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Answer
(i) We know that,
An Exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. ∠EDB is the exterior angle at vertex D. The interior opposite angle is ∠CAB.
⇒ ∠EBD = ∠CAB = 50°
Hence, the value of ∠EDB = 50°.
(ii) We know that,
⇒ OD = OB [radii of same circle]
From figure,
⇒ ∠AOD + ∠DOB = 180° [Linear pairs]
⇒ 140° + ∠DOB = 180°
⇒ ∠DOB = 180° - 140°
⇒ ∠DOB = 40°
In ΔODB,
By angle sum property of triangle,
⇒ ∠DOB + ∠ODB + ∠OBD = 180°
⇒ 40° + 2∠OBD = 180°
⇒ 2∠OBD = 180° - 40°
⇒ 2∠OBD = 140°
⇒ ∠OBD =
⇒ ∠OBD = 70°
From figure,
⇒ ∠OBD + ∠EBD = 180° [Linear pairs]
⇒ 70° + ∠EBD = 180°
⇒ ∠EBD = 180° - 70°
⇒ ∠EBD = 110°
Hence, the value of ∠EBD = 110°.
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